Fundamental Theorem Of Calculus (by Miki Sepala)

Fundamental Theorem of Calculus Basic Properties of Integrals Upper and Lower Estimates Intermediate Value Theorem for Integrals First Part of the Fundamental Theorem of Calculus Second Part of the Fundamental Theorem of Calculus Fundamental Theorem of Calculus

Index

FAQ

Basic Properties of Integrals Through this section we assume that all functions are continuous on a closed interval I = [a,b]. Below r is a real number, f and g are functions. Basic Properties of Integrals c

1

∫ f ( x ) dx = 0

3

a

b

c

b

∫ f ( x ) dx = ∫ f ( x ) dx + ∫ f ( x ) dx a

5

a

∫ f ( x ) dx = −∫ f ( x ) dx

2

c

b

b

a

4

c

b

b

b

a

a

a

b

b

a

a

∫ r f ( x ) dx = r ∫ f ( x ) dx

∫ ( f ( x ) + g ( x ) ) dx = ∫ f ( x ) dx + ∫ g ( x ) dx

These properties of integrals follow from the definition of integrals as limits of Riemann sums. Index

Mika Seppälä: Fundamental Theorems

FAQ

Upper and Lower Estimates Theorem 1

If f ( x ) ≤ g ( x ) ≤ h ( x ) ∀x ∈ [a, b ], b

b

b

a

a

a

∫ f ( x ) dx ≤ ∫ g ( x ) dx ≤ ∫ h ( x ) dx. Especially:

{

}

b

{

}

min g ( x ) x ∈ [a, b ] ( b − a ) ≤ ∫ g ( x ) dx ≤ max g ( x ) x ∈ [a, b ] ( b − a ) . a

The rectangle bounded from above by the red line is contained in the domain bounded by the graph of g.

Index

The rectangle bounded from above by the green line contains the domain bounded by the graph of g.

Mika Seppälä: Fundamental Theorems

FAQ

Intermediate Value Theorem for Integrals Theorem 2

∃ξ ∈ [a, b ] such that

b

∫ f ( x ) dx = f (ξ )( b − a ). a

Proof

{

By the previous theorem,

}

1 min f ( x ) x ∈ [a, b ] ≤ b−a

b

∫ f ( x ) dx ≤ max {f ( x ) x ∈ [a, b]} a

By the Intermediate Value Theorem for Continuous Functions,

1 ∃ξ ∈ [a, b ] such that f (ξ ) = b−a Index

b

∫ f ( x ) dx. a

This proves the theorem.

Mika Seppälä: Fundamental Theorems

FAQ

First Part of the Fundamental Theorem of Calculus First Part of the Fundamental Theorem of Calculus

The function F ( x ) =

x

∫ f ( t ) dt is differentiable for

x ∈ ( a, b )

a

and F′ ( x ) = f ( x ) for all x ∈ ( a, b ) . Proof

Let h ≠ 0.

(

F( x + h) − F( x )

)

h

1 = h

x +h

By the properties of integrals.

∫ f ( t ) dt x

1 f (ξ h ) ( ( x + h ) − x ) = f (ξ h ) where h By the Intermediate Value ξh is between x and x + h. Theorem for Integrals =

As h → 0, ξ h → x. Since f is continuous, lim f (ξ h ) = f ( x ) . h →0

Index

Mika Seppälä: Fundamental Theorems

FAQ

Second Part of the Fundamental Theorem of Calculus Second Part of the Fundamental Theorem of Calculus

Assume that F is an antiderivative of a continuous b

∫ f ( x ) dx = F ( b ) − F ( a ) .

function f. Then

a

Proof

By the First Fundamental Theorem of Calculus, the function x

G ( x ) = ∫ f ( t ) dt is an antiderivative of the function f. a

b

We have G ( a ) = 0 and ∫ f ( x ) dx = G ( b ) = G ( b ) − G ( a ) . a

If F is a general antiderivative of the function f, then F ( x ) = G ( x ) + C b

for some constant C. Hence F ( b ) − F ( a ) = G ( b ) − G ( a ) = ∫ f ( x ) dx. a

Index

Mika Seppälä: Fundamental Theorems

FAQ

Fundamental Theorem of Calculus We collect the previous two results into one theorem.

Fundamental Theorem of Calculus Assume that f is a continuous function. x

1. The function g ( x ) = ∫ f ( t ) dt is an antiderivative of f. b

a

2. Let F be an antiderivative of f. Then

∫ f ( x ) dx = F ( b ) − F ( a ) . a

Notation

F ( x ) ⎤⎦ a = F ( b ) − F ( a ) . Other common notations for b

the same quantity F ( x ) a and ⎡⎣F ( x ) ⎤⎦ a . b

b

b

We have ∫ f ( x ) dx = F ( x ) ⎤⎦ a . b

a

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Mika Seppälä: Fundamental Theorems

FAQ

Examples (1) x

Example

Let f ( x ) = ∫ e− t dt . Compute f ′ ( x ) . 2

0

Solution

The function to be integrated in the formula defining f is continuous. Hence f ′ ( x ) = e

− x2

by the Fundamental

Theorem of Calculus.

Index

Mika Seppälä: Fundamental Theorems

FAQ

Examples (2) Example

Let g ( x ) =

x2

∫ 0

sin ( t ) dt . Compute g′ ( x ) . t

Solution

Here one must first observe that the function h ( t ) =

sin ( t )

is

t everywhere continuous provided that we set h ( 0 ) = 1. Hence the integral is well defined and we can apply the Fundamental Theorem of Calculus. u sin ( t ) sin ( u ) 2 Let f ( u ) = ∫ dt , and u ( x ) = x . Then f ′ ( u ) = , t u 0 g ( x ) = f ( u ( x ) ) ⇒ g′ ( x ) = f ′ ( u ( x ) ) u′ ( x ) = Index

( ) ( 2x ) = x

sin x 2

Mika Seppälä: Fundamental Theorems

2

( )

2sin x 2 x FAQ